Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{2(z - 1)}{2z} \times \dfrac{5}{9z - 9} $
When multiplying fractions, we multiply the numerators and the denominators. $t = \dfrac{ 2(z - 1) \times 5 } { 2z \times (9z - 9) } $ $ t = \dfrac {5 \times 2(z - 1)} {2z \times 9(z - 1)} $ $ t = \dfrac{10(z - 1)}{18z(z - 1)} $ We can cancel the $z - 1$ so long as $z - 1 \neq 0$ Therefore $z \neq 1$ $t = \dfrac{10 \cancel{(z - 1})}{18z \cancel{(z - 1)}} = \dfrac{10}{18z} = \dfrac{5}{9z} $